LMF2013: Ejercicios de deducción natural en lógica de primer orden con Isabelle/HOL (2)
En la clase de hoy del curso Lógica matemática y fundamentos se han resuelto los ejercicios 22, 27, 28, 29, 32 y 34 de la relación 7 sobre deducción natural en lógica de primer orden con Isabelle/HOL.
Las soluciones de los ejercicios resueltos se muestran a continuación:
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text {* --------------------------------------------------------------- Ejercicio 22. Demostrar {∀x. P x ⟶ Q x ∨ R x, ¬(∃x. P x ∧ R x)} ⊢ ∀x. P x ⟶ Q x ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_22a: "⟦∀x. P x ⟶ Q x ∨ R x; ¬(∃x. P x ∧ R x)⟧ ⟹ ∀x. P x ⟶ Q x" by auto -- "La demostración estructurada es" lemma ejercicio_22b: assumes "∀x. P x ⟶ Q x ∨ R x" "¬(∃x. P x ∧ R x)" shows "∀x. P x ⟶ Q x" proof fix a show "P a ⟶ Q a" proof assume "P a" have "P a ⟶ Q a ∨ R a" using assms(1) .. hence "Q a ∨ R a" using `P a` .. thus "Q a" proof assume "Q a" thus "Q a" . next assume "R a" with `P a` have "P a ∧ R a" .. hence "∃x. P x ∧ R x" .. with assms(2) show "Q a" .. qed qed qed -- "La demostración detallada es" lemma ejercicio_22c: assumes "∀x. P x ⟶ Q x ∨ R x" "¬(∃x. P x ∧ R x)" shows "∀x. P x ⟶ Q x" proof (rule allI) fix a show "P a ⟶ Q a" proof (rule impI) assume "P a" have "P a ⟶ Q a ∨ R a" using assms(1) by (rule allE) hence "Q a ∨ R a" using `P a` by (rule mp) thus "Q a" proof (rule disjE) assume "Q a" thus "Q a" by this next assume "R a" with `P a` have "P a ∧ R a" by (rule conjI) hence "∃x. P x ∧ R x" by (rule exI) with assms(2) show "Q a" by (rule notE) qed qed qed text {* --------------------------------------------------------------- Ejercicio 27. Demostrar o refutar ((∀x. P x) ∨ (∀x. Q x)) ⟷ (∀x. P x ∨ Q x) ------------------------------------------------------------------ *} lemma ejercicio_27: "((∀x. P x) ∨ (∀x. Q x)) ⟷ (∀x. P x ∨ Q x)" oops (* Auto Quickcheck found a counterexample: P = {a\<^isub>1} Q = {a\<^isub>2} *) text {* --------------------------------------------------------------- Ejercicio 28. Demostrar o refutar ((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x) ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_28a: "((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x)" by auto -- "La demostración estructurada es" lemma ejercicio_28b: "((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x)" proof assume "(∃x. P x) ∨ (∃x. Q x)" thus "∃x. P x ∨ Q x" proof assume "∃x. P x" then obtain a where "P a" .. hence "P a ∨ Q a" .. thus "∃x. P x ∨ Q x" .. next assume "∃x. Q x" then obtain a where "Q a" .. hence "P a ∨ Q a" .. thus "∃x. P x ∨ Q x" .. qed next assume "∃x. P x ∨ Q x" then obtain a where "P a ∨ Q a" .. thus "(∃x. P x) ∨ (∃x. Q x)" proof assume "P a" hence "∃x. P x" .. thus "(∃x. P x) ∨ (∃x. Q x)" .. next assume "Q a" hence "∃x. Q x" .. thus "(∃x. P x) ∨ (∃x. Q x)" .. qed qed -- "La demostración detallada es" lemma ejercicio_28c: "((∃x. P x) ∨ (∃x. Q x)) ⟷ (∃x. P x ∨ Q x)" proof (rule iffI) assume "(∃x. P x) ∨ (∃x. Q x)" thus "∃x. P x ∨ Q x" proof (rule disjE) assume "∃x. P x" then obtain a where "P a" by (rule exE) hence "P a ∨ Q a" by (rule disjI1) thus "∃x. P x ∨ Q x" by (rule exI) next assume "∃x. Q x" then obtain a where "Q a" by (rule exE) hence "P a ∨ Q a" by (rule disjI2) thus "∃x. P x ∨ Q x" by (rule exI) qed next assume "∃x. P x ∨ Q x" then obtain a where "P a ∨ Q a" by (rule exE) thus "(∃x. P x) ∨ (∃x. Q x)" proof (rule disjE) assume "P a" hence "∃x. P x" by (rule exI) thus "(∃x. P x) ∨ (∃x. Q x)" by (rule disjI1) next assume "Q a" hence "∃x. Q x" by (rule exI) thus "(∃x. P x) ∨ (∃x. Q x)" by (rule disjI2) qed qed text {* --------------------------------------------------------------- Ejercicio 29. Demostrar o refutar (∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y) ------------------------------------------------------------------ *} lemma ejercicio_29: "(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)" quickcheck (* Quickcheck found a counterexample: P = (λx. undefined)(a\<^isub> := {b}, b := {a}) *) oops text {* --------------------------------------------------------------- Ejercicio 32. Demostrar o refutar ∃x y. R x y ∨ R y x; ¬(∃x. R x x)⟧ ⟹ ∃x y. x ≠ y ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_32a: fixes R :: "'c ⇒ 'c ⇒ bool" assumes "∃x y. R x y ∨ R y x" "¬(∃x. R x x)" shows "∃(x::'c) y. x ≠ y" using assms by metis -- "La demostración estructurada es" lemma ejercicio_32b: fixes R :: "'c ⇒ 'c ⇒ bool" assumes "∃x y. R x y ∨ R y x" "¬(∃x. R x x)" shows "∃(x::'c) y. x ≠ y" proof - obtain a where "∃y. R a y ∨ R y a" using assms(1) .. then obtain b where "R a b ∨ R b a" .. hence "a ≠ b" proof assume "R a b" show "a ≠ b" proof assume "a = b" hence "R b b" using `R a b` by (rule subst) hence "∃x. R x x" .. with assms(2) show False .. qed next assume "R b a" show "a ≠ b" proof assume "a = b" hence "R a a" using `R b a` by (rule ssubst) hence "∃x. R x x" .. with assms(2) show False .. qed qed hence "∃y. a ≠ y" .. thus "∃(x::'c) y. x ≠ y" .. qed -- "La demostración detallada es" lemma ejercicio_32c: fixes R :: "'c ⇒ 'c ⇒ bool" assumes "∃x y. R x y ∨ R y x" "¬(∃x. R x x)" shows "∃(x::'c) y. x ≠ y" proof - obtain a where "∃y. R a y ∨ R y a" using assms(1) by (rule exE) then obtain b where "R a b ∨ R b a" by (rule exE) hence "a ≠ b" proof (rule disjE) assume "R a b" show "a ≠ b" proof (rule notI) assume "a = b" hence "R b b" using `R a b` by (rule subst) hence "∃x. R x x" by (rule exI) with assms(2) show False by (rule notE) qed next assume "R b a" show "a ≠ b" proof (rule notI) assume "a = b" hence "R a a" using `R b a` by (rule ssubst) hence "∃x. R x x" by (rule exI) with assms(2) show False by (rule notE) qed qed hence "∃y. a ≠ y" by (rule exI) thus "∃(x::'c) y. x ≠ y" by (rule exI) qed text {* --------------------------------------------------------------- Ejercicio 34. Demostrar o refutar {∀x. P a x x, ∀x y z. P x y z ⟶ P (f x) y (f z)⟧ ⊢ ∃z. P (f a) z (f (f a)) ------------------------------------------------------------------ *} -- "La demostración automática es" lemma ejercicio_34a: "⟦∀x. P a x x; ∀x y z. P x y z ⟶ P (f x) y (f z)⟧ ⟹ ∃z. P (f a) z (f (f a))" by metis -- "La demostración estructura es" lemma ejercicio_34b: assumes "∀x. P a x x" "∀x y z. P x y z ⟶ P (f x) y (f z)" shows "∃z. P (f a) z (f (f a))" proof - have "P a (f a) (f a)" using assms(1) .. have "∀y z. P a y z ⟶ P (f a) y (f z)" using assms(2) .. hence "∀z. P a (f a) z ⟶ P (f a) (f a) (f z)" .. hence "P a (f a) (f a) ⟶ P (f a) (f a) (f (f a))" .. hence "P (f a) (f a) (f (f a))" using `P a (f a) (f a)` .. thus "∃z. P (f a) z (f (f a))" .. qed -- "La demostración detallada es" lemma ejercicio_34c: assumes "∀x. P a x x" "∀x y z. P x y z ⟶ P (f x) y (f z)" shows "∃z. P (f a) z (f (f a))" proof - have "P a (f a) (f a)" using assms(1) by (rule allE) have "∀y z. P a y z ⟶ P (f a) y (f z)" using assms(2) by (rule allE) hence "∀z. P a (f a) z ⟶ P (f a) (f a) (f z)" by (rule allE) hence "P a (f a) (f a) ⟶ P (f a) (f a) (f (f a))" by (rule allE) hence "P (f a) (f a) (f (f a))" using `P a (f a) (f a)` by (rule mp) thus "∃z. P (f a) z (f (f a))" by (rule exI) qed |